Drawing vector diagrams

Question

The question is: Show that if $u$ and $v$ are vectors in $\mathbb{R^3}$, and v is non zero, that: $$w = u - \frac{u · v}{||v||^2}v$$ is orthogonal to $v$.

This part was fine until I tried to draw a diagram. If $w = u - nv$ (since $(u · v)/||v||^2$ is a constant) is it not parallel to $v$?

Answer 1

Note that

$$w\cdot v= \left(u - \frac{u · v}{||v||^2}v\right)\cdot v=u\cdot v - \frac{u · v}{||v||^2}v\cdot v=u\cdot v-u\cdot v=0$$

For the diagram refer to the general scheeme for Gram–Schmidt process with $u=v_2$ and $w=u_2$

Answer 2

Let us regroup the equation a bit: $$ w = u - \frac{u \cdot v}{\lVert v \rVert^2} v = u - \left( u \cdot \frac{v}{\lVert v \rVert} \right) \frac{v}{\lVert v \rVert} = u - \underbrace{(u \cdot e_v) \, e_v}_{u_v} $$

where $e_v$ is the unit vector in $v$-direction.

This $w$ is $u$ minus that part of $u$, which points into direction of $v$, so what is left over is orthogonal to $v$:

\begin{align} w \cdot v &= \left( u - (u \cdot e_v) \, e_v \right) \cdot v \\ &= u \cdot v - (u \cdot e_v) (e_v \cdot v) \\ &= u \cdot v - (\lVert u \rVert e_u \cdot e_v)(e_v \cdot \lVert v \rVert e_v) \\ &= u \cdot v - \lVert u \rVert \lVert v \rVert (e_u \cdot e_v) \\ &= 0 \end{align}

You will see this decomposition again for the Gram-Schmidt orthogonalisation method.